[英]How to properly serialize/deserialize rust?
我有多個結構對應於僅在運行時才知道的序列化/反序列化對象,例如:
#[derive(Serialize, Deserialize)]
struct Car{
model: i32,
year: i32
}
#[derive(Serialize, Deserialize)]
struct Person{
name: String,
age: i32
}
然后我有序列化和反序列化的功能:
fn deserialize(data: Vec<u8>){
let msg = str::from_utf8(data);
serde_json::from_str(msg);
}
fn serialize(&self, object: Car) -> String{
let msg = serde_json::to_string(&object).unwrap();
return msg;
}
如何使反序列化 function 反序列化為 Car 和 Person(可能還有許多其他不同類型)並返回 object? 我怎樣才能讓序列化 function 做同樣的事情:序列化 Car、Person 和其他對象(在函數的屬性中接受這些類型)?
您可以在Deserialize
特征上使deserialize
化泛型:
fn deserialize<'a, T: Deserialize<'a>>(data: &'a [u8]) -> T {
let msg = str::from_utf8(data).unwrap();
serde_json::from_str(msg).unwrap()
}
請注意,您需要一些生命周期,因為某些類型需要從反序列化字符串doc中借用。
您也可以使serialize
通用:
fn serialize<T: Serialize>(object: &T) -> String {
serde_json::to_string(object).unwrap()
}
您想使用泛型函數來允許傳入不同的類型,並設置特征邊界以確保對象能夠被序列化/反序列化。 調用serialize
時,會根據參數的類型推斷類型,但調用deserialize
時,如果無法推斷,則需要使用 turbofish ( ::<>
) 指定類型。
use serde::{Serialize, Deserialize};
use std::str;
#[derive(Serialize, Deserialize)]
struct Car {
model: i32,
year: i32
}
#[derive(Serialize, Deserialize)]
struct Person {
name: String,
age: i32
}
// constrain output types to have the `Deserialize` trait
fn deserialize<'a, T>(data: &'a [u8]) -> T where T: Deserialize<'a> {
let msg = str::from_utf8(data).unwrap();
serde_json::from_str::<T>(msg).unwrap()
}
// shorthand for the above when `T` isn't needed in the function body
fn serialize(object: &impl Serialize) -> String {
let msg = serde_json::to_string(object).unwrap();
return msg;
}
fn main() {
let car = Car { model: 7, year: 2077 };
let person = Person { name: "Bob".to_string(), age: 42 };
// types are infrerred from the parameters
let car_json = serialize(&car);
let person_json = serialize(&person);
let _: Car = deserialize(car_json.as_bytes()); // output type can be inferred
let _ = deserialize::<Car>(car_json.as_bytes()); // requres turbofish
let _: Person = deserialize(person_json.as_bytes()); // works for `Person` too
}
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